First Order Circuits
Objective of Lecture
 Explain the operation of a RC circuit in dc circuits
 As the capacitor releases energy when there is:


a transition in a unit step function voltage or current source
or a voltage or current source is switched out of the circuit.
 Explain the operation of a RL circuit in dc circuit
 As the inductor releases energy when there is:


a transition in a unit step function voltage or current source
or a voltage or current source is switched out of the circuit.
Natural Response
 The behavior of the circuit with no external sources of
excitation.
 There was a transition in the source in the circuit where
the unit step function changed from 1 to 0 at t ≤ 0s.
 There is stored energy in the capacitor or inductor at time
t = 0 s.
 For t > 0 s, the stored energy is released


Current flows through the circuit and voltages exist across
components in the circuit as the stored energy is released.
The stored energy will decays to zero as time approaches
infinite, at which point the currents and voltages in the circuit
become zero.
RC Circuit
 Suppose there is some charge on a capacitor at time t =
0s. This charge could have been stored because a
voltage or current source had been in the circuit at
t<0s, but was switched off at t = 0s.
 We can use the equations relating voltage and current
to determine how the charge on the capacitor is
removed as a function of time.
 The charge flows from one plate of the capacitor
through the resistor R to the other plate to neutralize the
charge on the opposite plate of the capacitor.
Equations for RC Circuit
 vC  vR  0
iC  iR
dvC
iC  C
dt
vR
iR 
R
C
dVC VR

0
dt
R
VR  VC
dVC VC

0
dt
RC
1 dVC
1

0
VC dt
RC
dVC
1

dt
VC
RC
ln VC   

t
 ln VC
RC
t t o

2
If Vo  VC
VC (t )  Vo e
t 0 s

and   RC
t

t
when t  0s
Vo
I R (t )   I C (t )  e
R

t

Since the voltages are equal and the
currents have the opposite sign, the
power that is dissipated by the resistor
is the power that is being released by
the capacitor.
2t
Vo  
p R (t )  VR I R 
e
R
CVo
w(t )   pR (t )dt 
2
0s
2
2t
 


1  e 


RL Circuits
VL  VR  0
IL  IR
dI L
VL  L
dt
I R  VR R
dI L
L
 RI R  0
dt
dI L RI L

0
dt
L
1 dI L R
 0
I L dt L
dI L
R
  dt
IL
L

R
ln I L    t  ln I L t 0 s
L

If I o  I L t 0 s
L
and  
R
2t

2

p R (t )  VR I R  RI o e
t
I L (t )  I o e

t

when t  0 s
VR (t )  VL (t )  RI o e

t

Since the voltages are equal and the
currents have the opposite sign, the
power that is dissipated by the resistor
is the power that is being released by
the capacitor.
LI o
w(t )   p R (t )dt 
2
0s
2
2t
 


1  e 


Initial Condition
 Can be obtained by inserting a d.c. source to the
circuit for a time much longer than  at least t = -5.
 Capacitor
 Vo is the voltage calculated by replacing the capacitor with a
resistor with infinite resistance (an open circuit) after the
voltage across the capacitor has reached a constant value
(steady state).
 Inductor
 Io is the current flowing through the inductor calculated by
replacing the inductor with a resistor with zero resistance (a
short circuit) after the current flowing through the inductor
has reached a constant value (steady-state).
PSpice
 You can set the initial condition on a capacitor or
inductor by doubling clicking on the part symbol. Then,
enter a value for IC in the pop-up window that opens.
Time constant, 
 The time required for the voltage across the capacitor
or current in the inductor to decay by a factor of 1/e or
36.8% of its initial value.
Example #1
IL(t)
V(t) = 6V [1 - u(t)]
Example #1 (con’t)
IL(t)
Example #1 (con’t)
Find the initial condition.
Io
t < 0s
VL = 0V
VR = 6V
IL = IR = 2mA
Therefore,
Io = 2mA
Example #1 (con’t)
IL(t)
Example #1 (con’t)
IL(t)
t > 0s
t = L/R = 10mH/3kW/ = 3.33ms
IL = IR =Ioe-t/ = 2mA e-(t/3.33ms)
VR = 3kW IR = 6V e-(t/3.33ms)
VL = L dIL/dt = -6V e-(t/3.33ms)
Note VR + VL = 0 V
Example #2
+
VC
_
Example #2 (con’t)
+
VC
_
Example #2 (con’t)
 Calculate the initial condition - the voltage on the
capacitor. Replace the capacitor with an open circuit
and find the voltage across the two terminals.
 Note that in this circuit, current will flow through R3 so there will
be a voltage across C, but it will not be equal to the magnitude of
the voltage source in the circuit.
Example #2 (con’t)
 The voltage across the capacitor is equal to the voltage
across the 12kW resistor.
VC = Vo = [12kW /15kW] 5V = 4V
Example #2 (con’t)
+
VC
_
+
VC
_
Example #2 (con’t)
 Further simplification of the circuit
+
VC
_
where Req  1kW  3kW 12kW  3kW
Example #2 (con’t)
  Req C  3kW(2 mF )  6ms
when t  2ms
VC  VC (t ) t  2 ms e
 ( t  2 ms ) / 
IR(t)
VC (t )  4Ve ( t  2 ms ) / 6 ms
VC (t )  VReq (t )
dVC (t )
 2 mF (4V / 6ms)e ( t  2 ms ) / 6 ms
dt
I C (t )  1.33mA e ( t  2 ms ) / 6 ms
I C (t )  C
I Req (t )   I C (t )  1.33mA e ( t  2 ms ) / 6 ms
I Req (t )  I C (t )  0
IC(t)
Summary
 The initial condition for:
 the capacitor voltage (Vo) is determined by replacing the
capacitor with an open circuit and then calculating the
voltage across the terminals.
 The inductor current (Io) is determined by replacing the
inductor with a short circuit and then calculating the current
flowing through the short.
 The time constant for:
 an RC circuit is   RC
 an RL circuit is   L/R
 The general equations for the natural response of:
 the voltage across a capacitor is
VC (t )  Vo e  (t t ) /
 the current through an inductor is I (t )  I e (t t ) /
o
o
L
o